Ir 235tw

Infrared Spectroscopy

• Spectroscopy is an instrumentally aided studies of the interactions between matter (sample being analyzed) and energy (any portion of the electromagnetic spectrum, EMS) • EMS refers to the seemingly diverse collection of radiant energy, from cosmic rays to X-rays to visible light to microwaves, each of which can be considered as a wave or particle traveling at the speed of light. • Chemists can use portions of the EMS to selectively manipulate the energies contained within a molecule, to uncover detailed evidence of its chemical structure and bonding.

EMS and Molecular Effects

=>

• Energy (E)

E = hn = hc/l

– where h is Planck’s constant, c is the speed of light, n is frequency or the number of vibrations per second and l is the wavelength

• Wavenumber (n’) – given in cm-1

• Period (P)

= hcn’

n’ = 1/ l

P = 1/n

– the time between a vibration

 Energy, frequency, and wavenumber are directly proportional to each other.

 The four most common spectroscopic methods used in organic analysis are:

Method

Ultraviolet-Visible Spectroscopy Infrared Spectroscopy Nuclear Magnetic Resonance Mass Spectroscopy

Abbrev.

Energy used UV-Vis ultravioletvisible IR infrared NMR MS

radio frequencies electron volts

Units

nm mm or cm-1 Hz amu

What actually happens to the sample during an analysis? {How do the sample and energy ―interact‖?}

Matter/Energy Interactions • What happens when a sample absorbs UV/Vis energy? excitation of ground state electrons (typically p and n electrons) Eelectronic increases momentarily

p* UV/Vis (200 nm)

p  p*

sample

transition

p • What happens when a sample absorbs IR energy? stretching and bending of bonds (typically covalent bonds) Evibration increases momentarily

-O-H

IR (3500 cm-1)

- O —H

• Infrared spectroscopy (IR) measures the bond vibration frequencies in a molecule and is used to determine the functional group • Mass spectrometry (MS) fragments molecule and measures the masses

the

• Nuclear magnetic resonance (NMR) spectroscopy detects signals from hydrogen atoms and can be used to distinguish isomers

• Ultraviolet (UV) spectroscopy uses electron transitions to determine bonding patterns

The IR Region • Just below red in the visible region • Wavelengths usually 2.5-25 mm

• More common units are wavenumbers, or cm-1, the reciprocal of the wavelength in centimeters (104/mm = 4000-400 cm-1) • Wavenumbers are proportional to frequency and energy • The IR region is divided into three regions: the near, mid, and far IR. The mid IR region is of greatest practical use to the organic chemist.

Molecular Vibrations and IR Spectroscopy

Molecules are made up of atoms linked by chemical bonds. The movement of atoms and chemical bonds like spring and balls (vibration)

Vibrations

What is a vibration in a molecule? Any change in shape of the molecule- stretching of bonds, bending of bonds, or internal rotation around single bonds

There are two main vibrational modes : 1. Stretching - change in bond length (higher frequency)

Stretching vibration

Stretching Types

Symmetric

Asymmetric

2. Bending - change in bond angle (lower frequency) Bending Types

In-plane (Scissoring)

Out-plane (Twisting)

Modes of vibrations Stretching: change in bond distance. Occurs at higher energy: 4000-1250 cm1. Bending: change in bond angle. Occurs at lower energy: 1400-666 cm1.

H 2O

-CH2-

 More complex types of stretching and bending are possible

 Can a vibration change the dipole moment of a molecule?

 Asymmetrical stretching/bending and internal rotation change the dipole moment of a molecule. Asymmetrical stretching/bending are IR active.

 Symmetrical stretching/bending does not. Not IR active  Infrared active vibrations (those that absorb IR radiation) must result in a change of dipole

moment

Fundamental Vibrations (Absorption Frequencies) A molecule has as many as degrees of freedom as the total degree of freedom of its individual atoms. • Each atom has 3 degree of freedom (x,y,z) • A molecule of n atoms therefore has 3n degrees of freedom. • Non linear molecules (e.g. H2O) Vibrational degrees of freedom or Fundamental Vibrations = 3n – 6 H

H O

Symmetrical Stretching (υs OH) 3652 cm-1

H

H O

Asymmetrical Stretching (υas OH) 3756 cm-1

H

H O

Scissoring (δs HOH) 1596 cm-1

• For linear molecule (e.g. CO2) : Vibrational degrees of freedom or Fundamental Vibrations = 3n – 5 O

C

O

Symmetrical Stretching (υs CO2) 1340 cm-1

O

C

O

Scissoring (bending out

of the plane of the paper)

(δs CO2) 666 cm-1

O

C

O

Asymmetrical Stretching (υas CO2) 2350 cm-1

O

C

O

Scissoring (bending in the plane of the paper) (δs CO2) 666 cm-1

―BACKGROUND‖

Carbon Dioxide and Water

H-O-H

H-O-H O=C=O

 The theoretical no. of fundamental vibrations will seldom be observed because overtones (multiples of a given frequencies) and combination tones (sum of two other vibrations) increase the no. of bands.

 Other phenomena reduce the no. of bands including: 1. Fundamental frequencies that fall outside the 4000-400 cm-1 region.

2. Fundamental bands that are too weak to be observed. 3. Fundamental bands that are so close that they coalesce. 4. The occurrence of a degenerate band from several absorptions of the same frequency in highly symmetrical molecules. 5. The failure of certain fundamental vibrations to appear in the IR because of the lack of change in molecular dipole.

Which of the following atoms or molecules will

absorb IR radiation: I—Cl

H2

N2

Why?

Cl2

IR-Active and Inactive • A polar bond is usually IR-active. • A nonpolar bond in a symmetrical molecule will absorb weakly or not at all.

How does the mass influence the vibration? H2

MM =2 g/mole

I2

MM =254 g/mole

The greater the mass - the lower the wavenumber (ύ)

For a vibration at 4111 cm-1 (the stretch in H2), how many vibrations occur in a second? 120 trillion vibration per second!!!!

120 x 1012 vibrations/sec or a vibration every 8 x 10-15 seconds!

Hooke’s Law

M1

ύ = The vibration frequency (cm-1) c = Velocity of light (cm/s) f = force constant of bond (dyne/cm)

M1 and M2 are mass (g) of atom M1 and M2

M2

 The relative contributions of bond strength is also considered in vibrational frequencies.  In general functional groups that have a strong dipole give rise to strong absorptions in the IR.

THE EQUATION OF A SIMPLE HARMONIC OSCILLATOR

n= c =

frequency in cm-1 velocity of light

( 3 x 1010 cm/sec ) 1

n = 2pc

K

m

K = force constant in dynes/cm

where

m

m1 m2 = m1 + m2

C C > C C >C C multiple bonds have higher K’s

m = atomic masses This equation describes the vibrations of a bond.

m = reduced mass

larger K, higher frequency

1

n = 2pc

K

m

larger atom masses, lower frequency increasing K

constants

= C=C > C=C > C-C 2150

1650

1200

increasing m

C-H > C-C > C-O > C-Cl > C-Br 3000

1200

1100

750

650

Stretching Frequencies

• Frequency decreases with increasing atomic mass. • Frequency increases with increasing bond energy.

EXAMPLE: Calculate the fundamental frequency expected in the infrared absorption spectrum for the C - O stretching frequency. The value of the force constant is 5.0 X 105 dynes/cm.

wave number = ύ 1 ύ= ------- * ((k(m1 + m2))/(m1 m2))1/2 2pc 1 ύ = ---------------------- cm-1 2(3.14)(3.0E10) * ((5.0E5(12+16)(6.02E23))/(12*16))1/2 ύ = 1112 cm-1

50% 0%

Transmission

100%

Vibrational Spectroscopy

4500

4000

3500

3000

Wavenumber

2500 (cm-1)

2000

1500

IR Correlation Diagram Region I 3600-2700 cm-1

Transmittance (%)

100 80 60

O-H N-H C-H bond stretching

C=O Fingerprint Region

acid chlorides anhydrides (below 1500 cm-1) esters ketones aldehydes carboxylic acids amides

alcohols phenols carboxylic acids

40

amines amides

20 0 4000 2.5

Region II 1800-1600 cm-1

alkynes alkenes alkanes

3500

3000 3.0

C-H =C-H -C-H

2500 4.0

2000 5.0

1500 6.0

1000 10.0

Frequency (cm-1) / Wavelength (microns, mm)

Instrumentation 1. Radiation source 3. Solvents, sample cells, samples

2. Monochromator 4. Readout / Recorder

schematic diagram of a double beam double-grating infrared spectrophotometer

SOLVENTS, CELLS, SAMPLES

Solvents 1. Must be transparent in the region studied: no single solvent is transparent throughout the entire IR region 2. Water and alcohols are seldom employed to avoid O-H band of water . 3. Must be chemically inert (does not react with substance or cell holder). CCl4, CS2, or CHCl3; may be used but we should consider its IR spectrum

Cells - NaCl or KCl cells may be used (moisture from air and sample should be avoided: even with care, their surfaces eventually become fogged

due to absorption of moisture) - Very thin (path length = 0.1 to 1.0 mm) - Sample concentration = about 0.1 – 10%

Samples 1. Solid KBr disk (1 mg solid sample + 100 mg KBr pressed into a disk) Mull: 1 mg solid sample suspended in Nujol (heavy liquid hydrocarbon)

2. Liquid Neat (thin film of liquid between two NaCl plates solution in CCl4 and put in special NaCl cells.

3. Gas IR spectrum is obtained directly by permitting the sample to expand into an evacuated special cells.

Interferogram  The interferogram at the right displays the interference pattern and contains all of the spectrum information.  A Fourier transform converts the time domain to the frequency domain with absorption as a function of frequency.

Use of IR spectra • Identification of functional groups on a molecule – this is a very important tool in organic chemistry • Spectral matching can be done by computer software and library spectra • Since absorbance follows Beer’s Law, can do quantitative analysis

FEATURES OF AN IR SPECTRUM • An IR spectrum is a plot of per cent transmittance (or absorbance) against wavenumber (frequency or wavelength). A typical infrared spectrum is shown below. • A 100 per cent transmittance in the spectrum implies no absorption of IR radiation. When a compound absorbs IR radiation, the intensity of transmitted radiation decreases. This results in a decrease of per cent transmittance and hence a dip in the spectrum. The dip is often called an absorption peak or absorption band. • Different types of groups of atoms (C-H, O-H, N-H, etc…) absorb infrared radiation at different characteristic wavenumbers.

IR Spectrum

Baseline

Absorbance/ Peak

• No two molecules will give exactly the same IR spectrum (except enantiomers) • Simple stretching: 1600-3500 cm-1 • Complex vibrations: 400-1400 cm-1, called the ―fingerprint region‖

Infrared Spectrum of Propanal (CH3CH2CHO) 100 90

Per cent transmittance

80 70 60 50 40

30 20 10

C=O 0 4600

3800

3000

2200

1800

1400

Wavenumber (cm-1)

1000

800

600

400

Describing IR Absorptions IR absorptions are described by their frequency and appearance. • Frequency (n) is given in wavenumbers (cm-1) • Appearance is qualitative: intensity and shape • conventional abbreviations: vs

very strong

s m w br sh

strong medium weak broad sharp OR shoulder

• In general, the IR spectrum can be split into four regions for interpretation: • 4000  2500 cm-1: Absorption of single bonds formed by hydrogen and other elements e.g. OH, NH, CH • 2500  2000 cm-1: Absorption of triple bonds e.g. C≡C, C≡N • 2000  1500 cm-1: Absorption bonds e.g. C=C, C=O

of

double

• 1500  400 cm-1: This region often consists of many different, complicated bands. This part of the spectrum is unique to each compound and is often called the fingerprint region. It is rarely used for identification of particular functional groups.

Summary of IR Absorptions

BASE VALUES (+/- 10 cm-1)

O-H N-H C-H

3600 3400 3000

C N C C

2250 2150

C=O C=C

1715 1650

C O

~1100

These are the minimum number of values to memorize.

large range

O-H STRETCH

Typical Infrared Absorption Regions WAVELENGTH (mm)

O-H 2.5

4

O-H

C-H

5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

6.1

C=O

C N

N-H

5.5

2000

1800

FREQUENCY (cm-1)

6.5

15.4

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1650

1550

650

The O-H stretching region • O-H • O-H

3600 cm-1 (alcohol, free) 3300 cm-1 (alcohols & acids, H-bonding)

broadens

shifts FREE

3600

H-BONDED

3300

HYDROGEN-BONDED HYDROXYL R H

Many kinds of OH bonds of different lengths and strengths This leads to a broad absorption.

R

O

H O H R O

R O

H

H R

O

H

O R

“Neat” solution.

Longer bonds are weaker and lead to lower frequency.

Hydrogen bonding occurs in concentrated solutions ( for instance, undiluted alcohol ).

―FREE‖ HYDROXYL The ―free‖ hydroxyl vibrates without interference from any other molecule. Distinct bond has a welldefined length and strength.

CCl4 CCl4 R O

CCl4 H

CCl4

CCl4

Solvent molecules surround but do not hydrogen bond.

Occurs in dilute solutions of alcohol in an “inert” solvent like CCl4.

ALCOHOL

Cyclohexanol neat solution

, OH

O-H H-bond

CH2 C-O

C-H

CARBOXYLIC ACID

Butanoic Acid neat solution O-H H-bond

C-O CH2 C-H

C=O

O

CH3 CH2 CH2 C

OH

CARBOXYLIC ACID DIMER

O

H O C R

R C O H

O

Strong hydrogen bonding in the dimer weakens the OH bond and leads to a broad peak at lower frequency.

N-H STRETCH

Typical Infrared Absorption Regions N-H 2.5

WAVELENGTH (mm) 4

O-H

C-H

5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

2000

6.5

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1800 1650

FREQUENCY (cm-1)

15.4

1550

650

The N-H stretching region N-H

3300 - 3400 cm-1

• Primary amines give two peaks H

H

N

N H

symmetric

asymmetric

H

• Secondary amines give one peak • Tertiary amines give no peak

PRIMARY AMINE aliphatic

1-Butanamine

NH2 scissor NH2

CH3

CH2

CH3 CH2 CH2 CH2 NH2

PRIMARY AMINE aromatic

3-Methylbenzenamine

-CH3 Ar-H

NH2

NH2

benzene CH3

Ar-H

SECONDARY AMINE

N -Ethylbenzenamine

NH NH CH2 CH3

CH3 benzene

Ar-H

TERTIARY AMINE

N,N -Dimethylaniline Ar-H -CH3

no N-H

CH3 N CH3

CH3

benzene

Ar-H

C-H STRETCH

Typical Infrared Absorption Regions 2.5

C-H O-H

WAVELENGTH (mm) 4

C-H

5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S)

4000

2500

2000

15.4

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1800 1650

FREQUENCY (cm-1) We will look at this area first

6.5

1550

650

The C-H stretching region BASE VALUE = 3000 cm-1

•C-H sp stretch ~ 3300 cm-1 •C-H sp2 stretch > 3000 cm-1

UNSATURATED

3000 divides

•C-H sp3 stretch < 3000 cm-1

SATURATED

•C-H aldehyde, two peaks (both weak) ~ 2850 and 2750 cm-1

STRONGER BONDS HAVE LARGER FORCE CONSTANTS AND ABSORB AT HIGHER FREQUENCIES increasing frequency (cm-1)

3300 = =C-H

3100

3000 =C-H

2900 -C-H

2850 2750 -CH=O (weak)

sp-1s

sp2-1s

sp3-1s

aldehyde

increasing s character in bond increasing CH Bond Strength increasing force constant K CH BASE VALUE = 3000 cm-1

ALKANE

Hexane CH bending vibrations discussed shortly

CH stretching vibrations

includes CH3 sym and asym CH2 sym and asym

CH3 CH2 CH2 CH2 CH2 CH3

C-H BENDING

THE C-H BENDING REGION • CH2 bending ~ 1465 cm-1

• CH3 bending (asym) appears near the CH2 value ~ 1460 cm-1 • CH3 bending (sym) ~ 1375 cm-1

METHYLENE GROUP BENDING VIBRATIONS Scissoring

H

~1465 cm-1

C

Wagging

H

H

H C

~1250 cm-1

H

~1250 cm-1

C

C H

H

H ~720 cm-1

H C

H

H C

H Rocking

Twisting

in-plane

out-of-plane

Bending Vibrations

METHYLENE AND METHYL BENDING VIBRATIONS CH2

CH3 C-H Bending, look near 1465 and 1375 cm-1

1465

asym

sym

1460

1375

these two peaks frequently overlap and are not resolved

C H

H H

METHYLENE AND METHYL BENDING VIBRATIONS ADDITIONAL DETAILS FOR SYM CH3

CH2 asym

The sym methyl peak splits when you have more than one CH3 attached to a carbon.

CH3 sym

C CH3

1465

1460

geminal dimethyl (isopropyl) t-butyl

one peak

1375

1380 1390

CH3

1370

C

1370

CH3

CH3

two peaks

C CH3 CH3

two peaks

ALKANE

Hexane CH2 rocking CH2 bend

CH3 bend

CH stretch

CH3 CH2 CH2 CH2 CH2 CH3

ALKENE

1-Hexene

=CH

C=C

CH

CH2 CH3 bend CH

CH2 CH CH2 CH2 CH2 CH3

AROMATIC

Toluene

Ar-H

CH3

CH3

C=C benzene Ar-H

ALKYNE

1-Hexyne

= C=C CH2, CH3 = =C-H

C-H

HC

C

CH2 CH2 CH2 CH3

Fingerprinting

C10H22

C12H26

Similar But Not Identical

C N AND C C STRETCH

Typical Infrared Absorption Regions C=N = 2.5

4

O-H

C-H

C=C =

WAVELENGTH (mm) 5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

2000

6.5

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1800 1650

FREQUENCY (cm-1)

15.4

1550

650

The triple bond stretching region • C =N • C =C

2250 cm-1 2150 cm-1

The cyano group often gives a strong, sharp peak due to its large dipole moment. The carbon-carbon triple bond gives a sharp peak, but it is often weak due to a lack of a dipole. This is especially true if it is at the center of a symmetric molecule. R C C R

NITRILE

Propanenitrile

BASE = 2250

= C=N

CH3 CH2 C

N

ALKYNE BASE = 2150

1-Hexyne

= =C-H

= C=C

HC

C

CH2 CH2 CH2 CH3

C=O STRETCHING

Typical Infrared Absorption Regions C=O

WAVELENGTH (mm) 2.5

4

O-H

C-H

5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

2000

6.5

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1800 1650

FREQUENCY (cm-1)

15.4

1550

650

THE CARBONYL STRETCHING REGION  This region stretches from about 1800 to 1650 cm-1 - RIGHT IN THE MIDDLE OF THE SPECTRUM  The base value is 1715 cm-1 (ketone)  The bands are very strong !!! due to the large C=O dipole moment.  C=O is often one of the strongest peaks in the spectrum

KETONE BASE = 1715

2-Butanone 1715

overtone 2x C=O

C-H

CH bend O

C=O C=O CH3 C

CH2 CH3

C=O IS SENSITIVE TO ITS ENVIRONMENT EACH DIFFERENT KIND OF C=O COMES AT A DIFFERENT FREQUENCY acid chloride

ester

aldehyde

ketone

carboxylic acid

amide

O

O

O

O

O

O

R C

R C

R C

R C

R C

R C

Cl

1800

OR'

1735

1725

H

R

1715

OH

1710

NH2

1690

anhydride

O

O

R C O C R

1810 and 1760 ( two peaks )

BASE VALUE THESE VALUES ARE WORTH LEARNING all are +/- 10 cm-1

C=O BOND LENGTHS IN CARBONYL COMPOUNDS shorter

longer

1.225 A

1.231 A

1.235 A

1.248 A

acid chloride

ester

ketone

amide

1780 cm-1

1735 cm-1

1715 cm-1

1680 cm-1

SUMMARY Ketones are at lower frequency than Aldehydes because of the second electron-donating alkyl group.

Acid chlorides are at higher frequency than ketones because of the electron-withdrawing halide.

Esters are at higher frequencies than ketones due to the electron-withdrawing oxygen atom. This is more important than resonance with the electron pair on the oxygen.

Amides

are at lower frequencies than ketones due to resonance involving the unshared pair on nitrogen. The electronwithdrawing effect of nitrogen is less important than the resonance. Note the electronegativity difference, O versus N, weights the two factors (resonance/ e-withdrawal) differently in esters than in amides.

Acids are at lower frequency than ketones due to H-bonding.

KETONE

overtone of strong C=O peak 1719 x 2 = 3438

BASE = 1715

2-Butanone overtone

C-H

CH bend O

C=O CH3 C

3438

CH2 CH3

ALDEHYDE BASE = 1725

Nonanal CHO

CH bend C=O O CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 C

3460

H

ACID CHLORIDE

Dodecanoyl Chloride

BASE = 1800

CH bend O

C-H

3608

C=O

C CH3 (CH2)10

Cl

ESTER

Ethyl Butanoate

BASE = 1735

C-O

C-H

O

C=O

3482

C CH3 CH2 CH2

O

CH2 CH3

CARBOXYLIC ACID BASE = 1710

2-Methylpropanoic Acid O-H

O

C-H

C=O

C-O

C CH3 CH CH3

OH

CARBOXYLIC ACID DIMER RECALL

lowers frequency of C=O

O

H O C R

R C O H

and also of O-H

O

Strong hydrogen-bonding in the dimer weakens the O-H and C=O bonds and leads to broad peaks at lower frequencies.

AMIDE two peaks sym / asym

Propanamide

BASE = 1690

C-H

O

C=O NH2

CH bend

C CH3 CH2

NH2

ALKENES AROMATICS

C=C STRETCHING

Typical Infrared Absorption Regions C=C

WAVELENGTH (mm) 2.5

4

O-H

C-H

5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

2000

6.5

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1800 1650

FREQUENCY (cm-1)

15.4

1550

650

The C=C stretching region • C=C double bond at 1650 cm-1 is often weak or not even seen. • C=C benzene ring shows peak(s) near 1600 and 1400 cm-1 , one or two at each value CONJUGATION LOWERS THE VALUE. • When C=C is conjugated with C=O it is stronger and comes at a lower frequency.

ALKENE

1-Hexene

=C-H

C=C C-H aliphatic

CH2

C-H bend oops CH CH2 CH2 CH2 CH3

AROMATIC

Toluene

benzene C-H

CH3

benzene C=C oops

Typical Infrared Absorption Regions C-O

WAVELENGTH (mm) 2.5

4

O-H

C-H

5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

2000

6.5

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1800 1650

FREQUENCY (cm-1)

15.4

1550

650

C-O STRETCHING

The C-O stretching region • The C-O band appears in the range of 1300 to 1000 cm-1

• Look for one or more strong bands appearing in this range! • Ethers, alcohols, esters and carboxylic acids have C-O bands

Dibutyl Ether

C-H

ETHER BASE = 1100

CH2 CH3 bending

CH3 CH2 CH2 CH2 O

C-O CH2 CH2 CH2 CH3

AROMATIC ETHER BASE = 1100

Anisole

C-H aromatic O

CH3

oops

benzene C-O

ALCOHOL BASE = 3600 BASE = 1100

Cyclohexanol CH2 bend OH

OH

C-O C-H

CARBOXYLIC ACID

2-Methylpropanoic Acid

OH O

CH

C=O

C-O

C CH3 CH CH3

OH

ESTER

Ethyl Butanoate

CH C-O C=O

O C

CH3 CH2 CH2

O

CH2 CH3

N=O STRETCHING

Typical Infrared Absorption Regions WAVELENGTH (mm) 2.5

4

O-H

C-H

5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

2000

N-O

1800 1650

FREQUENCY (cm-1)

6.5

15.4

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1550

650

The N=O stretching region • N=O stretching -- 1550 and 1350 cm-1 asymmetric and symmetric stretchings

• Often the 1550 cm-1 peak is stronger than the other one

NITROALKANE

2-Nitropropane

C-H N=O NO2

gem-dimethyl

CH CH3

CH3

N=O

Typical Infrared Absorption Regions C-Cl

WAVELENGTH (mm) 2.5

4

O-H

C-H

5

6.1

C=O

C N

N-H

5.5

C C X=C=Y

Very few bands

(C,O,N,S) 4000

2500

2000

6.5

C-Cl C-O C=C C-N C-C N=O N=O * C=N

1800 1650

FREQUENCY (cm-1)

15.4

1550

650

The C-X stretching region • C-Cl 785 to 540 cm-1, often hard to find amongst the fingerprint bands!!

• C-Br and C-I appear outside the useful range of infrared spectroscopy. • C-F bonds can be found easily, but are not that common.

Often used as a solvent for IR spectra. When it is used, spectra show C-Cl absorptions.

Carbon Tetrachloride

Cl Cl

C Cl

Cl

C-Cl

Chlorobenzene

benzene ring combination bands

Cl

oops

benzene C=C

C-Cl

=C-H OUT OF PLANE BENDING

OUT-OF-PLANE BENDING (OOPS)

PLANE H ALKENES

above

H H

H

also with BENZENES

below

ALKENES Monosubstituted Disubstituted

10 R

11

13

s

H

H

R

R

s

H

H

H

R C C

R R

s H H

C C

1,1R

H

R

R

Trisubstituted

s

m

C C R

H

R

R C C

Tetrasubstituted R

15m

s

C C

trans-1,2-

14

H C C

cis-1,2-

12

R

=C-H OUT OF PLANE BENDING 1000

900

800

700 cm-1

BENZENES

10

11

12

13

14

Monosubstituted

s

15 m s

Disubstituted s ortho meta

m

s

para

s

RING H’s OOPS

Trisubstituted m

1,2,4

s

1,2,3

s

1,3,5 combination bands

s

s 1000

900

m m

800

700 cm-1